3.77 \(\int \frac{1}{(5+3 \cosh (c+d x))^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{45 \sinh (c+d x)}{512 d (3 \cosh (c+d x)+5)}-\frac{3 \sinh (c+d x)}{32 d (3 \cosh (c+d x)+5)^2}-\frac{59 \tanh ^{-1}\left (\frac{\sinh (c+d x)}{\cosh (c+d x)+3}\right )}{1024 d}+\frac{59 x}{2048} \]

[Out]

(59*x)/2048 - (59*ArcTanh[Sinh[c + d*x]/(3 + Cosh[c + d*x])])/(1024*d) - (3*Sinh[c + d*x])/(32*d*(5 + 3*Cosh[c
+ d*x])^2) - (45*Sinh[c + d*x])/(512*d*(5 + 3*Cosh[c + d*x]))

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Rubi [A]  time = 0.0631422, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2664, 2754, 12, 2657} \[ -\frac{45 \sinh (c+d x)}{512 d (3 \cosh (c+d x)+5)}-\frac{3 \sinh (c+d x)}{32 d (3 \cosh (c+d x)+5)^2}-\frac{59 \tanh ^{-1}\left (\frac{\sinh (c+d x)}{\cosh (c+d x)+3}\right )}{1024 d}+\frac{59 x}{2048} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Cosh[c + d*x])^(-3),x]

[Out]

(59*x)/2048 - (59*ArcTanh[Sinh[c + d*x]/(3 + Cosh[c + d*x])])/(1024*d) - (3*Sinh[c + d*x])/(32*d*(5 + 3*Cosh[c
+ d*x])^2) - (45*Sinh[c + d*x])/(512*d*(5 + 3*Cosh[c + d*x]))

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2657

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2*ArcTan[(b*Cos[c + d*x])/(a + q + b*Sin[c + d*x])])/(d*q), x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rubi steps

\begin{align*} \int \frac{1}{(5+3 \cosh (c+d x))^3} \, dx &=-\frac{3 \sinh (c+d x)}{32 d (5+3 \cosh (c+d x))^2}-\frac{1}{32} \int \frac{-10+3 \cosh (c+d x)}{(5+3 \cosh (c+d x))^2} \, dx\\ &=-\frac{3 \sinh (c+d x)}{32 d (5+3 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (5+3 \cosh (c+d x))}+\frac{1}{512} \int \frac{59}{5+3 \cosh (c+d x)} \, dx\\ &=-\frac{3 \sinh (c+d x)}{32 d (5+3 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (5+3 \cosh (c+d x))}+\frac{59}{512} \int \frac{1}{5+3 \cosh (c+d x)} \, dx\\ &=\frac{59 x}{2048}-\frac{59 \tanh ^{-1}\left (\frac{\sinh (c+d x)}{3+\cosh (c+d x)}\right )}{1024 d}-\frac{3 \sinh (c+d x)}{32 d (5+3 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (5+3 \cosh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.187694, size = 58, normalized size = 0.72 \[ \frac{59 \tanh ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )-\frac{3 (182 \sinh (c+d x)+45 \sinh (2 (c+d x)))}{(3 \cosh (c+d x)+5)^2}}{1024 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 3*Cosh[c + d*x])^(-3),x]

[Out]

(59*ArcTanh[Tanh[(c + d*x)/2]/2] - (3*(182*Sinh[c + d*x] + 45*Sinh[2*(c + d*x)]))/(5 + 3*Cosh[c + d*x])^2)/(10
24*d)

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Maple [A]  time = 0.017, size = 108, normalized size = 1.3 \begin{align*} -{\frac{9}{512\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) ^{-2}}+{\frac{69}{1024\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) ^{-1}}+{\frac{59}{2048\,d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) }+{\frac{9}{512\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) ^{-2}}+{\frac{69}{1024\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) ^{-1}}-{\frac{59}{2048\,d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5+3*cosh(d*x+c))^3,x)

[Out]

-9/512/d/(tanh(1/2*d*x+1/2*c)+2)^2+69/1024/d/(tanh(1/2*d*x+1/2*c)+2)+59/2048/d*ln(tanh(1/2*d*x+1/2*c)+2)+9/512
/d/(tanh(1/2*d*x+1/2*c)-2)^2+69/1024/d/(tanh(1/2*d*x+1/2*c)-2)-59/2048/d*ln(tanh(1/2*d*x+1/2*c)-2)

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Maxima [A]  time = 1.03263, size = 169, normalized size = 2.09 \begin{align*} -\frac{59 \, \log \left (3 \, e^{\left (-d x - c\right )} + 1\right )}{2048 \, d} + \frac{59 \, \log \left (e^{\left (-d x - c\right )} + 3\right )}{2048 \, d} - \frac{3 \,{\left (241 \, e^{\left (-d x - c\right )} + 295 \, e^{\left (-2 \, d x - 2 \, c\right )} + 59 \, e^{\left (-3 \, d x - 3 \, c\right )} + 45\right )}}{256 \, d{\left (60 \, e^{\left (-d x - c\right )} + 118 \, e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, e^{\left (-3 \, d x - 3 \, c\right )} + 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*cosh(d*x+c))^3,x, algorithm="maxima")

[Out]

-59/2048*log(3*e^(-d*x - c) + 1)/d + 59/2048*log(e^(-d*x - c) + 3)/d - 3/256*(241*e^(-d*x - c) + 295*e^(-2*d*x
- 2*c) + 59*e^(-3*d*x - 3*c) + 45)/(d*(60*e^(-d*x - c) + 118*e^(-2*d*x - 2*c) + 60*e^(-3*d*x - 3*c) + 9*e^(-4
*d*x - 4*c) + 9))

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Fricas [B]  time = 2.17252, size = 1661, normalized size = 20.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*cosh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2048*(1416*cosh(d*x + c)^3 + 1416*(3*cosh(d*x + c) + 5)*sinh(d*x + c)^2 + 1416*sinh(d*x + c)^3 + 7080*cosh(d
*x + c)^2 + 59*(9*cosh(d*x + c)^4 + 12*(3*cosh(d*x + c) + 5)*sinh(d*x + c)^3 + 9*sinh(d*x + c)^4 + 60*cosh(d*x
+ c)^3 + 2*(27*cosh(d*x + c)^2 + 90*cosh(d*x + c) + 59)*sinh(d*x + c)^2 + 118*cosh(d*x + c)^2 + 4*(9*cosh(d*x
+ c)^3 + 45*cosh(d*x + c)^2 + 59*cosh(d*x + c) + 15)*sinh(d*x + c) + 60*cosh(d*x + c) + 9)*log(3*cosh(d*x + c
) + 3*sinh(d*x + c) + 1) - 59*(9*cosh(d*x + c)^4 + 12*(3*cosh(d*x + c) + 5)*sinh(d*x + c)^3 + 9*sinh(d*x + c)^
4 + 60*cosh(d*x + c)^3 + 2*(27*cosh(d*x + c)^2 + 90*cosh(d*x + c) + 59)*sinh(d*x + c)^2 + 118*cosh(d*x + c)^2
+ 4*(9*cosh(d*x + c)^3 + 45*cosh(d*x + c)^2 + 59*cosh(d*x + c) + 15)*sinh(d*x + c) + 60*cosh(d*x + c) + 9)*log
(cosh(d*x + c) + sinh(d*x + c) + 3) + 24*(177*cosh(d*x + c)^2 + 590*cosh(d*x + c) + 241)*sinh(d*x + c) + 5784*
cosh(d*x + c) + 1080)/(9*d*cosh(d*x + c)^4 + 9*d*sinh(d*x + c)^4 + 60*d*cosh(d*x + c)^3 + 12*(3*d*cosh(d*x + c
) + 5*d)*sinh(d*x + c)^3 + 118*d*cosh(d*x + c)^2 + 2*(27*d*cosh(d*x + c)^2 + 90*d*cosh(d*x + c) + 59*d)*sinh(d
*x + c)^2 + 60*d*cosh(d*x + c) + 4*(9*d*cosh(d*x + c)^3 + 45*d*cosh(d*x + c)^2 + 59*d*cosh(d*x + c) + 15*d)*si
nh(d*x + c) + 9*d)

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Sympy [A]  time = 5.73531, size = 445, normalized size = 5.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*cosh(d*x+c))**3,x)

[Out]

Piecewise((-59*log(tanh(c/2 + d*x/2) - 2)*tanh(c/2 + d*x/2)**4/(2048*d*tanh(c/2 + d*x/2)**4 - 16384*d*tanh(c/2
+ d*x/2)**2 + 32768*d) + 472*log(tanh(c/2 + d*x/2) - 2)*tanh(c/2 + d*x/2)**2/(2048*d*tanh(c/2 + d*x/2)**4 - 1
6384*d*tanh(c/2 + d*x/2)**2 + 32768*d) - 944*log(tanh(c/2 + d*x/2) - 2)/(2048*d*tanh(c/2 + d*x/2)**4 - 16384*d
*tanh(c/2 + d*x/2)**2 + 32768*d) + 59*log(tanh(c/2 + d*x/2) + 2)*tanh(c/2 + d*x/2)**4/(2048*d*tanh(c/2 + d*x/2
)**4 - 16384*d*tanh(c/2 + d*x/2)**2 + 32768*d) - 472*log(tanh(c/2 + d*x/2) + 2)*tanh(c/2 + d*x/2)**2/(2048*d*t
anh(c/2 + d*x/2)**4 - 16384*d*tanh(c/2 + d*x/2)**2 + 32768*d) + 944*log(tanh(c/2 + d*x/2) + 2)/(2048*d*tanh(c/
2 + d*x/2)**4 - 16384*d*tanh(c/2 + d*x/2)**2 + 32768*d) + 276*tanh(c/2 + d*x/2)**3/(2048*d*tanh(c/2 + d*x/2)**
4 - 16384*d*tanh(c/2 + d*x/2)**2 + 32768*d) - 816*tanh(c/2 + d*x/2)/(2048*d*tanh(c/2 + d*x/2)**4 - 16384*d*tan
h(c/2 + d*x/2)**2 + 32768*d), Ne(d, 0)), (x/(3*cosh(c) + 5)**3, True))

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Giac [A]  time = 1.15086, size = 123, normalized size = 1.52 \begin{align*} \frac{59 \, \log \left (3 \, e^{\left (d x + c\right )} + 1\right )}{2048 \, d} - \frac{59 \, \log \left (e^{\left (d x + c\right )} + 3\right )}{2048 \, d} + \frac{3 \,{\left (59 \, e^{\left (3 \, d x + 3 \, c\right )} + 295 \, e^{\left (2 \, d x + 2 \, c\right )} + 241 \, e^{\left (d x + c\right )} + 45\right )}}{256 \, d{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} + 10 \, e^{\left (d x + c\right )} + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*cosh(d*x+c))^3,x, algorithm="giac")

[Out]

59/2048*log(3*e^(d*x + c) + 1)/d - 59/2048*log(e^(d*x + c) + 3)/d + 3/256*(59*e^(3*d*x + 3*c) + 295*e^(2*d*x +
2*c) + 241*e^(d*x + c) + 45)/(d*(3*e^(2*d*x + 2*c) + 10*e^(d*x + c) + 3)^2)